Prove the identity $ \sum\limits_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0.$

Question

Prove the identity $$ \sum_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0 $$

I want to reduce $$ \frac{(a+k-2)!}{(a+k-n)!}=(2-n)\binom{a+k-n}{a+k-2}^{-1}=(2-n) (a+k-2) \int_0^1 z^{a+k-2}(1-z)^{2-n} dz, $$ and then try integrate it but the binomial coefficient $$ \binom{a+k-n}{a+k-2}=\frac{(a+k-n)!}{(2-n)! (a+k-2)!}, $$ is undefined for $n>2$ however the sum seems correct for any $n.$

Any ideas?

Answer 1

We obtain for integral $n>0$ and $a\in\mathbb{C}\setminus\{n-1,n-2,n-3,\ldots\}$ \begin{align*} \color{blue}{\sum_{k=0}^{n-1}}&\color{blue}{(-1)^k\binom{n}{k}(n-k)\frac{(a+k-2)!}{(a+k-n)!}}\\ &=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k\frac{(a+n-k-2)!}{(a-k)!}\tag{1}\\ &=(-1)^nn(n-2)!\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\binom{a+n-k-2}{a-k}\tag{2}\\ &=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{a+n-k-3}{a-k-1}\tag{3}\\ &=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{-n+1}{a-k+1}\tag{4}\\ &=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}[z^{a-k+1}](1+z)^{-n+1}\tag{5}\\ &=(-1)^{n+1}n(n-2)^{-n+1}\sum_{k=0}^{n-1}\binom{n-1}{k}z^k\tag{6}\\ &=(-1)^{n+1}n(n-2)^{-n+1}(1+z)^{n-1}\tag{7}\\ &=(-1)^{n+1}n(n-2)![z^{a+1}]1\\ &\,\,\color{blue}{=0} \end{align*} and the claim follows.

Comment:

• In (1) we set the upper limit of the sum to $n$ (i.e. adding zero) and change the order of summation by $k\to n-k$.

• In (2) we apply the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ and use binomial coefficients.

• In (3) we shift the index to start with $k=0$.

• In (4) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^{q}$.

• In (5) we use the coefficient of operator $[z^p]$ to denote the coefficient of a series. This way we can write $[z^p](1+z)^q=\binom{q}{p}$.

• In (6) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

• In (7) we apply the binomial theorem.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &{1 \over \pars{n - 2}!}\bbox[10px,#ffd]{\ds{\sum_{k = 0}^{n - 1}\pars{-1}^{k}{n \choose k}\pars{n - k}{\pars{a + k - 2}! \over \pars{a + k - n}!}}} = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k} {a + k - 2 \choose a + k - n} \\[5mm] = &\ \sum_{k = 0}^{n}\pars{-1}^{n - k}{n \choose k}k {a + n - k - 2 \choose a - k} = \sum_{k = 0}^{n}\pars{-1}^{n - k}{n \choose k} {-n + 1\choose a - k}\pars{-1}^{a - k}\, k \\[5mm] = &\ \pars{-1}^{n + a}\sum_{k = 0}^{n}{n \choose k}k \bracks{z^{a - k}}\pars{1 + z}^{-n + 1} = \pars{-1}^{n + a}\bracks{z^{a}}\pars{1 + z}^{-n + 1} \sum_{k = 0}^{n}{n \choose k}k\,z^{k} \\[5mm] = &\ \pars{-1}^{n + a}\bracks{z^{a}}\pars{1 + z}^{-n + 1}\, z\partiald{}{z}\sum_{k = 0}^{n}{n \choose k}z^{k} = \pars{-1}^{n + a}\bracks{z^{a - 1}}\pars{1 + z}^{-n + 1}\, \partiald{\pars{1 + z}^{n}}{z} \\[5mm] = &\ \pars{-1}^{n + a}\bracks{z^{a - 1}}\pars{1 + z}^{-n + 1}\, \bracks{n\pars{1 + z}^{n - 1}} = \pars{-1}^{n + a}\, n\bracks{z^{a - 1}}1 = \pars{-1}^{n + 1}\, n\bracks{a = 1} \end{align}

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$$ \bbx{\bbox[10px,#ffd]{\ds{\sum_{k = 0}^{n - 1}\pars{-1}^{k}{n \choose k}\pars{n - k}{\pars{a + k - 2}! \over \pars{a + k - n}!}}} = \bracks{a = 1}\pars{-1}^{n + 1}\,n\pars{n - 2}!} $$